Question: Solve for $X$. $\dfrac{1}{3}X+\left[\begin{array}{rr}0 & 4 \\ 4 & 7 \\18 &6 \end{array}\right]=\left[\begin{array}{rr}4 & 0 \\ 0 & 1 \\-13 &15 \end{array}\right] $ $X=$
Solution: The Strategy First, we can represent the matrices of the equation with letters, which will make the equation easier to handle. Then we can solve the equation for $X$ and obtain an expression with the letters we defined. Finally, we can substitute back the actual matrices into the resulting expression, multiply the matrices through by the scalars, and simplify the result. Solving the equation for $X$ We are given the following equation. $\dfrac{1}{3}X+\left[\begin{array}{rr}0 & 4 \\ 4 & 7 \\18 &6 \end{array}\right]=\left[\begin{array}{rr}4 & 0 \\ 0 & 1 \\-13 &15 \end{array}\right] $ Let's represent the above matrices as follows. $A=\left[\begin{array}{rr}0 & 4 \\ 4 & 7 \\18 &6 \end{array}\right] ~~~~~~~~~ B = \left[\begin{array}{rr}4 & 0 \\ 0 & 1 \\-13 &15 \end{array}\right] $ Then we can rewrite the equation as follows. $\begin{aligned}\dfrac{1}{3}X+A&=B \\\dfrac{1}{3}X&=B-A \\X&=3(B-A)\end{aligned}$ Finding $X$ $\begin{aligned}X&=3(B-A) \\\\&=3\left(\left[\begin{array}{rr}4 & 0 \\ 0 & 1 \\-13 &15 \end{array}\right]-\left[\begin{array}{rr}0 & 4 \\ 4 & 7 \\18 &6 \end{array}\right] \right) \\\\\\&=3\left[\begin{array}{rr}4 & -4 \\ -4 & -6 \\-31 &9\end{array}\right] \\\\\\&=\left[\begin{array}{rr}12 & -12 \\ -12 & -18 \\-93 &27 \end{array}\right]\end{aligned}$ Summary $X=\left[\begin{array}{rr}12 & -12 \\ -12 & -18 \\-93&27 \end{array}\right]$